Answer:
9.1 = basic 1.2= very acidic 5.7= acidic
Explanation:
Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
Answer:
positively charged elctrons
1. 2 H2 + O2 = 2 H2O
2. 6 K + B2O3 = 3 K2O + 2 B
3. 10 Na + 2 NaNO3 = 6 Na2O + N2
Answer:
3
Explanation:
the three elements involved in this compound are Li, S, O.
lithium, sulfur, and oxygen. Which create the ionic compound, "Lithium Sulfate."
7 atoms total, since there are two lithium, four oxygen, and one sulfate atom. this is a white inorganic salt.
